6
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In how many ways can '$6$' things be distributed equally among $2$ groups ?

I tried

$\dbinom{6}{3}\times \dbinom{3}{3}\times 3!\times 3!$

But I am not sure if it is correct .

I look for a short and simple way.

I have studied maths up to $12$th grade.

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  • $\begingroup$ Why have you multiplied by 3! and 3! ? $\endgroup$ – Shailesh Sep 12 '15 at 1:25
  • $\begingroup$ To permute $3$ objects. $\endgroup$ – R K Sep 12 '15 at 1:28
  • $\begingroup$ Ncmathsadist has already answered the question. $\endgroup$ – Shailesh Sep 12 '15 at 1:29
  • $\begingroup$ Why scare quotes around 6? Is is not really 6 after all? $\endgroup$ – Henning Makholm Sep 12 '15 at 1:30
  • $\begingroup$ actually it was $'3\times 2'$. $\endgroup$ – R K Sep 12 '15 at 1:31
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It's $6\choose 3$ if you have groups $A$ and $B$. You choose three items and give them to $A$; there are $6\choose 3$ ways to do this. You then give the remaining to group $B$.

The $3!s$ are called for if you stipulate the order in which the groups receive the items is important. Otherwise, not.

$${6\choose 3} = 20.$$

If the two groups are indistinguishable and you are just interested in partitioning the items into two parts of size three, we have an overcount of a factor of 2. In that case there are only $10$ ways. You must be careful about exactly what you are counting.

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